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Data Interpretation and Probability Flashcards

6 cards from real BMST practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. A quality control inspector records the number of defective units in six consecutive batches as follows: 4, 7, 3, 7, 5, 8. What is the median number of defective units?

    Answer: 6

    To find the median, the data set must first be ordered from smallest to largest: 3, 4, 5, 7, 7, 8. Since there is an even number of data points (6), the median is the average of the two middle numbers (the 3rd and 4th values), which are 5 and 7. The average is (5 + 7) / 2 = 6.

  2. A toolbox contains 12 screws, 8 bolts, and 5 nails. If one item is drawn at random, what is the probability that it is a bolt?

    Answer: 8/25

    The probability of an event is the ratio of favorable outcomes to the total number of possible outcomes. The number of favorable outcomes (drawing a bolt) is 8. The total number of items is 12 (screws) + 8 (bolts) + 5 (nails) = 25. Therefore, the probability of drawing a bolt is 8/25.

  3. A line graph tracks the pressure inside a sealed container over a 6-hour period. The data points are as follows: Hour 0 (100 kPa), Hour 1 (105 kPa), Hour 2 (115 kPa), Hour 3 (130 kPa), Hour 4 (135 kPa). During which one-hour interval did the pressure increase the most?

    Answer: Hour 2 to Hour 3

    To find the largest increase, calculate the change in pressure for each interval: Hour 0-1 changed by 5 kPa (105-100). Hour 1-2 changed by 10 kPa (115-105). Hour 2-3 changed by 15 kPa (130-115). Hour 3-4 changed by 5 kPa (135-130). The greatest increase of 15 kPa occurred between Hour 2 and Hour 3.

  4. A system has two independent safety alarms. The probability of the first alarm failing is 0.05. The probability of the second alarm failing is 0.02. What is the probability that both alarms will fail simultaneously?

    Answer: 0.001

    For independent events, the probability of both occurring is found by multiplying their individual probabilities. Therefore, the probability of both alarms failing is 0.05 * 0.02 = 0.001.

  5. A technician recorded the following voltage readings: 12V, 15V, 11V, 15V, 13V, 10V, 15V. Which of the following statements is true for this data set?

    Answer: The mode is the largest value among the three measures.

    First, order the data: 10, 11, 12, 13, 15, 15, 15. The mode (most frequent value) is 15. The median (middle value) is 13. The mean (average) is (10+11+12+13+15+15+15) / 7 = 91 / 7 = 13. Comparing these values, the mode (15) is greater than both the mean (13) and the median (13), making it the largest.

  6. A laboratory uses a simple two-character identification code for its samples. The first character must be one of 4 specific letters (A, B, C, D) and the second character must be a single digit from 1 to 5, inclusive. How many unique identification codes are possible?

    Answer: 20

    This problem uses the fundamental counting principle. To find the total number of unique combinations, multiply the number of choices for each position. There are 4 choices for the first character and 5 choices for the second character. Total unique codes = 4 * 5 = 20.